Theo công thức tọa độ trung điểm:
\(\left\{{}\begin{matrix}\dfrac{x_A+x_B}{2}=x_P\\\dfrac{x_B+x_C}{2}=x_M\\\dfrac{x_C+x_A}{2}=x_N\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_A+x_B=2x_P=0\\x_B+x_C=2\\x_C+x_A=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_A=1\\x_B=-1\\x_C=3\end{matrix}\right.\)
Tương tự:
\(\left\{{}\begin{matrix}y_A+y_B=2y_P=-8\\y_B+y_C=2y_M=2\\y_A+y_C=2y_N=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y_A=-2\\y_B=-6\\y_C=8\end{matrix}\right.\)
Vậy: \(A\left(1;-2\right)\); \(B\left(-1;-6\right)\) ; \(C\left(3;8\right)\)
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